Problem: Is ${441702}$ divisible by $9$ ?
Answer: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {441702}= &&{4}\cdot100000+ \\&&{4}\cdot10000+ \\&&{1}\cdot1000+ \\&&{7}\cdot100+ \\&&{0}\cdot10+ \\&&{2}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {441702}= &&{4}(99999+1)+ \\&&{4}(9999+1)+ \\&&{1}(999+1)+ \\&&{7}(99+1)+ \\&&{0}(9+1)+ \\&&{2} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {441702}= &&\gray{4\cdot99999}+ \\&&\gray{4\cdot9999}+ \\&&\gray{1\cdot999}+ \\&&\gray{7\cdot99}+ \\&&\gray{0\cdot9}+ \\&& {4}+{4}+{1}+{7}+{0}+{2} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${441702}$ is divisible by $9$ if ${ 4}+{4}+{1}+{7}+{0}+{2}$ is divisible by $9$ Add the digits of ${441702}$ $ {4}+{4}+{1}+{7}+{0}+{2} = {18} $ If ${18}$ is divisible by $9$ , then ${441702}$ must also be divisible by $9$ ${18}$ is divisible by $9$, therefore ${441702}$ must also be divisible by $9$.